Mathematics of Physics and Engineering

Non-Rigid Systems of Points 71

where Vj(t) = rj(t) - rcM(t). Hence, (2.2.13) becomes

L 0 = M rCM x rCM + LCM, (2.2.15)

that is, the angular momentum of the system relative to a point O is equal to
the angular momentum of the center of mass relative to that point plus the
angular momentum of the system relative to the center of mass. Once again,
we see that the center of mass plays a very special role in the description
of the motion of a system of points.
We emphasize that LQ ^ Mr CM X rcM as long as LCM ^ 0. Note
also that the vector functions Xj (t) and ij (t) depend on the choice of the
reference frame.
Let us now compute the time derivative of LcM{t) using the differen-
tiation rules of vector calculus (1.3.3), (1.3.4), (1.3.6) (see page 26). Since
ij x ij = 0, we have
j n n n
—LCM = y~}rni xi x •»' "" 5ZmJ *•» x ^' = 5ZmJ xi x XJ' (2-2.16)
3 = 1 j=l j = l

To express dLcM/dt in terms of the forces Fj acting on the rrij, we would
like to use Newton's Second Law (2.1.1), and then we need the frame to be
inertial. The frame at the center of mass is usually not inertial, because
the center of mass can have a non-zero acceleration relative to an inertial
frame. Accordingly, we choose a convenient inertial frame O and apply
(2.2.4), page 67, in that frame:

ij = jr-j - TCM = Fj/rrij - F/M,

rrij Xj x ij = Xj x Fj - (rrij/M) Xj x F.

By (2.2.16) above,

dLcM

dt

and then (2.2.12) implies

n 1 / " \

j=i \j=i j

dLcM = Y,XjXFj. (2.2.17)

dt

If the internal forces i^L are central, then, by (2.2.9), these forces do not

contribute to the total torque. Since Xj —Xk = Tj — TCM — (>*fc — I"CM) =